All AP Physics 1 Resources
Example Question #1 : Series And Parallel
A circuit has a resistor with a resistance of followed by three parallel branches, each holding a resistor with a resistance of . What is the total equivalent resitance of the circuit?
First, we need to calculate the equivalent resistance of the three resistors in parallel. To do this, we will use the following equation:
Now, to get the total equivalent resistance, we can simply add the two remaining values, since they are in series:
Example Question #2 : Series And Parallel
Consider the given circuit:
A voltage is applied across points A and B so that current flows from A, to R2, to B. What is the value of this voltage if the current through R2 is 4A?
First, we need to calculate the current flow through R2 without the extra voltage attached. We will need to calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.
Then, we can use Ohm's law to calculate the current through the circuit:
Now that we have the current, we can calculate the additional current that the new voltage contributes:
There is only one resistor (R2) in the path of the new voltage, so we can calculate what that voltage needs to be to deliver the new current:
Example Question #3 : Series And Parallel
What is the effective resistance of this DC circuit?
None of the other answers
First, let's remind ourselves that the effective resistance of resistors in a series is and the effective resistance of resistors in parallel is .
Start this problem by determining the effective resistance of resistors 2, 3, and 4:
(This is because these three resistors are in series.)
Now, the circuit can be simplified to the following:
Next, we will need to determine the effective resistance of resistors and 6:
Again, the circuit can be simplified:
From here, the effective resistance of the DC circuit can be determined by calculating the effective resistance of resistors , 1, and 5:
Example Question #4 : Series And Parallel
Two lightbulbs, one graded at and one graded at are connected in series to a battery. Which one will be brighter? What if they are connected in parallel?
There's not enough information to complete this problem
The first step to figuring out this problem is to figure out how resistances of light bulb correlate to the power rating. For a resistor, the power dissipated is:
Thus, there is an inverse relationship between the resistance of the lightbulb and the power rating.
The second step is to take a look at circuit elements in series and in parallel. In series, they share the same current; in parallel they share the same voltage. Thus, for the two lightbulbs in series, the one with the higher resistance (lower wattage) will be brighter, and for a parallel configuration the one with the lower resistance (higher wattage) will be brighter.
Example Question #5 : Series And Parallel
If we have 3 resistors in a series, with resistor 1 having a resistance of , resistor 2 having a resistance of , and resistor 3 having a resistance of , what is the equivalent resistance of the series?
The total resistance of resistors in a series is the sum of their individual resistances. In this case,
Example Question #72 : Circuits
You are presented with three resistors, each measure . What is the difference between the total resistance of the resistors combined in series, and the total resistance of the resistors combined in parallel?
Resistors in series:
Resistors in parallel:
Example Question #6 : Series And Parallel
What is the total resistance of three resistors, , , and , in parallel?
The equation for equivalent resistance for multiple resistors in parallel is:
Plug in known values and solve.
Notice that for resistors in parallel, the total resistance is never greater than the resistance of the smallest element.
Example Question #7 : Series And Parallel
A circuit is created using a battery and 3 identical resistors, as shown in the figure. Each of the resistors has a resistance of . If resistor is removed from the circuit, what will be the effect on the current through resistor ?
The current through will decrease
The current through will increase by a factor of four
The current through will increase by a factor of two
The current through will remain the same
Cannot be determined without knowing the resistivity of the wire
The current through will decrease
Since the resistors and form a parallel network, removing from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of and , the increased resistance leads to lower current in each of these resistors.
Example Question #75 : Circuits
Determine the total charge stored by a circuit with 2 identical parallel-plate capacitors in parallel with area , and a distance of between the parallel plates. Assume the space between the parallel plates is a vacuum. The circuit shows a voltage difference of 10V.
To determine total charge stored, we need to add up the capacitance of each capacitor(because they are capacitors in parallel) and multiply by the voltage difference. Recall that for capacitors,
For parallel plate capacitors:
Here, , which is the permittivity of empty space, is the dielectric constant, which is since there is only vacuum present, , which is the area of the parallel plates, and , which is the distant between the plates.
Plug in known values to solve for capacitance.
Each of the two capacitors has capacitance
Since the capacitors are in parallel, the total capacitance is the sum of each individual capacitance. The total capacitance in the circuit is given by:
Plug this value into our first equation and solve for the total charge stored.
, where is the total charge stored in the capacitor. Since
Example Question #8 : Series And Parallel
What is the total current flowing through a system with 2 resistors in parallel with resistances of and , and a battery with voltage difference of 10V?
First we need to determine the overall resistance of the circuit before we know how much current is flowing through. Since the resistors are in parallel, their resistances will add reciprocally:
where is the total resistance of the circuit.
Now that we've solved for , we know that the current flowing through the circuit can be found using Ohm's law: